Question

cos 4(theta)= 8cos(theta)^4- 8cos(theta)^2+1​

Answer 1

Given : cos 4(theta)= 8cos(theta)^4- 8cos(theta)^2+1

Some trigonometrical identities :

$\mathtt{1) {sin}^{2} \: x + {cos}^{2} \: x = 1 }\implies\mathtt{ {sin}^{2} \: x = 1 - {cos}^{2} \: x } \\ \\ \mathtt{2)sin(x + y) = sin \: x \: \times cos \: y + cos \: x \times sin \: y} \\ \\ \mathtt{3)cos(x + y) = cos \: x \times cos \: y + sin \: x \times sin \: y}$

So,

$\mathtt{cos(4x) =cos(2x + 2x) } \\ \\ \\ \implies\mathtt{cos(2x) \times cos(2x) - sin(2x) \times sin(2x)} \\ \\ \\ \implies\mathtt{(cos(2x)) {}^{2} - (sin(2x)) {}^{2} } \\ \\ \\ \implies\mathtt{(cos(x + x)) {}^{2} - (sin(x + x)) {}^{2} } \\ \\ \\ \implies\mathtt{(cos \: x \times cos \: x - sin \: a \times sin \: x) {}^{2} - (sin \: x \times cos \: x + sin \: x \times cos \: x) {}^{2} } \\ \\ \\ \implies\mathtt{(cos {}^{2} \: x - sin {}^{2} \: x) {}^{2} - (2sin \: x \times cos \: x) {}^{2} } \\ \\ \\ \implies\mathtt{(cos {}^{2} \: x - sin {}^{2} \: x) {}^{2} - 4sin {}^{2} \: x \times cos {}^{2} \: x } \\ \\ \\ \implies\mathtt{(cos {}^{2} \: x - (1 - cos {}^{2} \: x)) {}^{2} - 4(1 - cos {}^{2} \: x)cos {}^{2} \: x } \\ \\ \\ \implies\mathtt{(2cos {}^{2} \: x - 1) {}^{2} - 4cos {}^{2} \: x + cos {}^{4} \: x } \\ \\ \\ \implies\mathtt{4cos {}^{4} \: x - 4cos {}^{2} \: x1 - 4cos {}^{2} \: x + 4cos {}^{4} \: x } \\ \\ \\ \implies\mathtt{8cos {}^{4} \: x - 8cos {}^{2} \: x + 1}$